📚 Solving Quadratic Equations by Factorisation | 因式分解法解一元二次方程
Quadratic equations appear throughout IGCSE Mathematics. The method of factorisation is one of the fastest ways to solve them when the expression can be written as a product of two linear factors.
二次方程在 IGCSE 数学中十分常见。当表达式可以写成两个一次因式相乘时，因式分解法是求解二次方程最快的方法之一。
1. What Is a Quadratic Equation? | 什么是二次方程？
are constants and a ≠ 0. The highest power of x is 2, which is why the graph of a quadratic is a curve called a parabola.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。方程中 x 的最高次数是 2，因此二次函数的图像是一条称为抛物线的曲线。
ax² + bx + c = 0, a ≠ 0
2x² + 5x = 0 and x² = 9. The last example can be rearranged to x² − 9 = 0 before solving.
例如 x² − 4x + 3 = 0、2x² + 5x = 0 和 x² = 9。最后一个方程可以先化为 x² − 9 = 0 再求解。
2. Why Factorisation Works | 为什么因式分解可行
The key idea is the zero product property. If two numbers multiply to give zero, then at least one of them must be zero. This property lets us split a quadratic equation into two simple linear equations.
核心原理是零乘积性质：如果两个数相乘等于零，那么其中至少有一个数等于零。这个性质让我们能把一个二次方程拆成两个简单的一次方程。
If p × q = 0, then p = 0 or q = 0.
e, if (x + 2)(x + 3) = 0, then either x + 2 = 0 or x + 3 = 0. This gives x = −2 or x = −3.
例如，如果 (x + 2)(x + 3) = 0，那么 x + 2 = 0 或 x + 3 = 0，所以 x = −2 或 x = −3。
3. Setting the Equation to Zero | 将方程化为标准形式
rearrange the equation so that one side is 0. The usual standard form is ax² + bx + c = 0.
因式分解前，必须先整理方程，使一边为 0。通常把方程化为标准形式 ax² + bx + c = 0。
x² = 5x + 6 → x² − 5x − 6 = 0
multiply to −6 and add to −5. These numbers are −6 and 1, so x² − 5x − 6 = (x − 6)(x + 1).
然后分解左边的多项式。对于 x² − 5x − 6，要找两个数，它们相乘等于 −6，相加等于 −5。这两个数是 −6 和 1，所以 x² − 5x − 6 = (x − 6)(x + 1)。
(x − 6)(x + 1) = 0 → x = 6 or x = −1
4. Common Factor and Difference of Two Squares | 公因式与平方差
by taking out a common factor first. For example, 2x² + 4x = 0 has a common factor of 2x.
有些二次方程先提取公因式会更简单。例如，2x² + 4x = 0 有公因式 2x。
2x² + 4x = 0 → 2x(x + 2) = 0
So x = 0 or x = −2. The difference of two squares is another important pattern.
所以 x = 0 或 x = −2。平方差公式也是一个重要的变形技巧。
x² − a² = (x − a)(x + a)
For x² − 25 = 0, we write x² − 25 = (x − 5)(x +
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